Problem: N different natural numbers are given. If you select four different ones and set them as $ A $, $ B $, $ C $, $ D $, the following formula

$ \ Frac {A + B} {C --D} $



I want to find the maximum value of.




Given N different natural numbers, choose 4 different from them and create a program to find the maximum value of the above formula.



Input

The input is given in the following format.


N
a1 a2 ... aN


The number N (4 ≤ N ≤ 1000) of natural numbers is given in the first line. The value ai (1 ≤ ai ≤ 108) of each natural number is given in the second line. However, the same natural number does not appear more than once (ai ≠ aj for i ≠ j).

Output

Outputs the maximum value of the above formula as a real number for a given N natural numbers. However, the error must not exceed plus or minus 10-5.

Examples

Input

10
1 2 3 4 5 6 7 8 9 10


Output

19.00000


Input

5
22 100 42 3 86


Output

9.78947


Input

6
15 21 36 10 34 5


Output

18.00000


Input

4
100000 99999 8 1


Output

28571.285714

Solution (in CPP):

Solution:
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
#include<functional>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<numeric>
#include<limits>
#include<iterator>

#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define rep(i,n) for(int i=0; i<n; i++)
#define INF (1<<20)

using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ll, char> plc;

int main()
{
	int n;
	cin >> n;
	vector<int> a(n);
	double ans = 0;

	for (int i = 0; i < n; i++)cin >> a[i];

	sort(a.begin(), a.end());
	reverse(a.begin(), a.end());

	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j < n; j++) {
			int A = 0, B = 1;
			if (A == i) {
				if (A + 1 != j)A = A + 1;
				else A = A + 2;
			}
			if (B == j) {
				if (B + 1 != A)B = B + 1;
				else B = B + 2;
			}
			ans = max(ans,((double)a[A]+a[B])/(a[i]-a[j]));
		}
	}
	printf("%.10f\n",ans);
	return	0;
}